what exactly is a residue - what are its applications[complex analysis]

[trickae]

In fundamental of Complex analysis (http://www.amazon.com/Fundamentals-Analysis-Applications-Engineering-Mathematics/dp/0139078746/ref=sr_1_1/002-0650139-6816057?ie=UTF8&s=books&qid=1177661695&sr=8-1) - residues is introduced as an exercise at the end of a chapter and thats it! (or it may resurface in a later chapter),

My question is that saff and snider looks at it as the numerator of the partial fraction exapansion of a polynomail fraction.

But in Schaums series we have a nice little function like this:


a = lim 1/(k-1)! . (d^(k-1) /dz^(k-1)) {(z-a)^k f(z)}
z->a


where the term in red is the differential operator and the order is determined by k-1

so whats this used for? which method is right? why choose one method over the other? And what is it beside the sum of all the residues at the singularities = the integral of the function that contains it - i.e. f(z) ?

sorry if this is a silly question.

[HallsofIvy]

If you integrate zn on a closed curve around 0 what do you get?
More generally, if you integrate (z-z0)n on a close curve containing z0 what do you get?

That's easy to answer. Since (z-z0)n is analytic everywhere except at z0, we get the same for all such curves so we can look at the circle, of radius 1 around z0, taking z= z_0+ e^{i\theta} so that (z-z_0)^n= e^{ni\theta} and dz= ie^{i\theta}d\theta:
\int (z-z_0)^n dz= \int_0^{2\pi}ie^{(n+1)i\theta}d\theta
As long as n is not equal to -1, that is
\frac{-i}{n+1}e^{(n+1)i\theta}
evaluated at \theta= 0 and \theta= 2\pi. But e^{(n+1)i\theta} is 0 at both ends so the integral is 0.

If n= -1, then e^{(n+1)i\theta}= e^0= 1 so the integral is just
\int_0^{2\pi} d\theta= 2\pi i[/itex].

Now, suppose f(z) is a function having a "pole of order n" at z= z0. That means it can be written as a power series with powers of z down to -n: a "Laurent series".
[tex]f(z)= a_{-n}(z-z_0)^{-n}+ \cdot\cdot\cdot+ a_{-1}(z-z_0)^{-1}+ \cdot\cdot\cdot
Integrating that, term by term, on a contour containing z0, every term gives 0 except the (z-z0)-1 term. That term gives 2\pi i a_{-1} so the integral is that.

The "residue" of f(z) at z0 is precisely the coefficient of (z-z0)-1 in a Laurent series expansion of f(z). It can be found using a formula similar to the formula for Taylor's series coefficients- That's the formula Schaum's outline is giving you. If f(z) is analytic inside a countour except for some points at which it has poles, the integral of f around that contour is just 2\pi i times the sum of the residues at each of those poles.

[trickae]

thanks a tonne halls of ivy